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Problem with an equation involving natural numbers

  1. Jun 3, 2014 #1
    I'm having trouble solving the equation m2 - n2 = 707, where n and m are natural numbers.

    Because there are 2 variables, even though they are discrete, the obvious thing to do would be to use another equation to solve for one of the variables and then insert the new form to the original equation. The problem is that I don't have another equation to insert anything into. I guess I'm supposed to use the fact that m and n are natural numbers as a hint, but looking at their properties on Wikipedia didn't provide any insight into the subject.

    Any help would be appreciated.
     
  2. jcsd
  3. Jun 3, 2014 #2

    adjacent

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    How many solutions do you need?
     
  4. Jun 3, 2014 #3
    All of them. :)

    I should add that I tried inserting the equation into Wolfram Alpha, and it provided me with 2 possible combinations of values, which makes sense, since both m and n are raised to the second power. I'm just clueless as to how the values were calculated, since again the only way to find the values of 2 variables is if you also have at least 2 equations that involve these variables, as far as I know.
     
    Last edited: Jun 3, 2014
  5. Jun 3, 2014 #4
    I might have figured this one out, at least partially. I know you can factor m2 - n2 as (m+n)(m-n), and that 707 is the same thing as 7 * 101.

    If (m+n) = 7 and (m-n) = 101, maybe it is possible to solve this problem.
     
  6. Jun 3, 2014 #5

    adjacent

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    You should try that method first and see if it's correct. Find that values that way and plug it into the equation and see if it returns correct answers.
    ------
    One way is making n zero and calculating m. Then you can make m zero and calculate n.So you will get two solutions :wink:
     
    Last edited: Jun 3, 2014
  7. Jun 3, 2014 #6
    Ok, I solved it. Basically I had to form 2 groups of equations: one where (m+n) = 707 and (m-n) = 1, and another one where (m+n) = 101 and (m-n) = 7

    Simply solving these two sets of equations gave me these 2 sets of solutions:

    1) m = 354 and n = 353
    2) m = 54 and n = 47

    Both m and n had to be positive integers, so I couldn't just arbitrarily set the values of (m+n) and (m-n). Since there was a minus sign in one of the factors, it actually mattered whether I set (m+n) to be 101 instead of 7, for example.
     
  8. Jun 3, 2014 #7

    SammyS

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    Yes. that's excellent insight.

    However if m & n are natural numbers, which is larger, (m+n) or (m-n) ?

    7 and 101 are both prime, so there's only one other way to factor 707.


    Oh ! I see you have now solved it! Excellent !



    By the way, 707 is not a perfect square, so adjacent's suggestion regarding setting n=0, won't work.
     
  9. Jun 3, 2014 #8

    adjacent

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    ##m^2-n^2=707##
    ##0^2-n^2=707##
    ##-n^2=707## --> ##-n=\sqrt{707}## --> ##n=-\sqrt{707}##
    ##m=0##
    ##n=-\sqrt{707}##
    ---------------
    ##m^2-n^2=707##
    ##m^2-0^2=707##
    ##n=0##
    ##m=\sqrt{707}##


    I don't see anything wrong here
     
  10. Jun 3, 2014 #9

    CAF123

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    The solution set is restricted to the natural numbers.
     
  11. Jun 3, 2014 #10

    adjacent

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    Oh. I had some misconceptions about natural numbers. I thought they were real numbers :redface:
     
  12. Jun 3, 2014 #11

    Ray Vickson

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    You should: ##n^2 = -707## does not have a real solution (the square of a nonzero real number is always > 0). The two solutions of the equation ##n^2 = -707## are ##\pm i \sqrt{707} \doteq \pm 26.58947160 i##, where ##i = \sqrt{-1}##. So, not only are the two solutions not integers, they are not even real.
     
  13. Jun 3, 2014 #12

    adjacent

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    Ok. I have done some algebra mistake there. Sorry, I understand now. :)
     
  14. Jun 4, 2014 #13
    This problem was a great one, since it forced me to use factoring in a new way, to form 2 groups of equations out of a single equation.

    Sometimes it's the basics that really get you confused.
     
    Last edited: Jun 4, 2014
  15. Jun 4, 2014 #14

    adjacent

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    I have learned that method from your solution too. It's great :smile:
     
  16. Jun 4, 2014 #15

    verty

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    This question was a bit ad hoc, I don't like questions like this because they don't teach anything, it's all about trying things until they work. I mean, if one didn't know that 101 is a prime number, and who would, it is difficult.

    Number theory questions like this are very common in math competitions unfortunately.
     
  17. Jun 4, 2014 #16

    ehild

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    You only have to check the primes less than √101 that is 10, if they are divisors of 101.

    The sum of the digits is 2: 3 is not divisor of 101
    101 does not end with 0 or 5: 5 is not divisor.
    7 , it is not a divisor.
    101 is prime!

    ehild
     
  18. Jun 4, 2014 #17

    pasmith

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    I would posit that it's obvious (or should be obvious) that [itex]m^2 - n^2 = (m + n)(m - n)[/itex] and hence that factorizing 707 is necessary; 7 is clearly a prime factor, so the only thing to determine is whether 101 is prime. We only have to check those primes which are less than [itex]\sqrt{101} \in (10, 11)[/itex]. 101 is obviously not divisible by 2, 3, or 5, so the only candidate prime factor is 7. There are many ways of showing that 7 is not a factor of 101, including simple long division.
     
  19. Jun 4, 2014 #18

    verty

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    If it's obvious, I think it has almost no value as a question, there is no conceptual nicety, no elegance. Unless it is meant to be a test of how to factorize a number, I don't see any redeeming value in doing it.

    And if it is a test of how to factorize a number, the student will have learned a simple algorithm, test each prime in order for divisibility. Or perhaps they will have learned that they only need to check up to the square root of the number. But either way, why not just ask directly: factorize this number: 707.

    And I mean, any student who is just learning how to factorize a number is likely to be very young and a dry question like this is going to frustrate them.

    Edit: I'll try to speak more objectively in future, speaking to whom the question seems to be aimed at is, I see now, confusing.
     
    Last edited: Jun 4, 2014
  20. Jun 4, 2014 #19

    ehild

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    @verty: The problem is about factorizing m^2-n^2 and connecting that with factorizing a number. In my country, such problems are aimed for students about age of 14.

    ehild
     
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